Question: $ F = \left[\begin{array}{rrr}5 & 0 & -2 \\ 5 & -1 & 2\end{array}\right]$ $ C = \left[\begin{array}{rr}3 & 0 \\ 5 & 1 \\ 4 & 1\end{array}\right]$ What is $ F C$ ?
Solution: Because $ F$ has dimensions $(2\times3)$ and $ C$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F C = \left[\begin{array}{rrr}{5} & {0} & {-2} \\ {5} & {-1} & {2}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{0} \\ {5} & \color{#DF0030}{1} \\ {4} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{3}+{0}\cdot{5}+{-2}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{3}+{0}\cdot{5}+{-2}\cdot{4} & ? \\ {5}\cdot{3}+{-1}\cdot{5}+{2}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{3}+{0}\cdot{5}+{-2}\cdot{4} & {5}\cdot\color{#DF0030}{0}+{0}\cdot\color{#DF0030}{1}+{-2}\cdot\color{#DF0030}{1} \\ {5}\cdot{3}+{-1}\cdot{5}+{2}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{3}+{0}\cdot{5}+{-2}\cdot{4} & {5}\cdot\color{#DF0030}{0}+{0}\cdot\color{#DF0030}{1}+{-2}\cdot\color{#DF0030}{1} \\ {5}\cdot{3}+{-1}\cdot{5}+{2}\cdot{4} & {5}\cdot\color{#DF0030}{0}+{-1}\cdot\color{#DF0030}{1}+{2}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}7 & -2 \\ 18 & 1\end{array}\right] $